657. Robot Return to Origin

 

Initially, there is a Robot at position (0, 0). Given a sequence of its moves, judge if this robot makes a circle, which means it moves back to the original place.

The move sequence is represented by a string. And each move is represent by a character. The valid robot moves are R (Right), L(Left), U (Up) and D (down). The output should be true or false representing whether the robot makes a circle.

Example 1:

Input: "UD"
Output: true

 

Example 2:

Input: "LL"
Output: false

 

这道题让我们判断一个路径是否绕圈,就是说有多少个U,就得对应多少个D。同理,L和R的个数也得相等。这不就是之前那道Valid Parentheses的变种么,这次博主终于举一反三了!这比括号那题还要简单,因为括号至少还有三种,这里就水平和竖直两种。比较简单的方法就是使用两个计数器,如果是U,则cnt1自增1;如果是D,cnt1自减1。同理,如果是L,则cnt1自增1;如果是R,cnt1自减1。最后只要看cnt1和cnt2是否同时为0即可,参见代码如下:

 

解法一:

class Solution {
public:
    bool judgeCircle(string moves) {
        int cnt1 = 0, cnt2 = 0;
        for (char move : moves) {
            if (move == 'U') ++cnt1;
            else if (move == 'D') --cnt1;
            else if (move == 'L') ++cnt2;
            else if (move == 'R') --cnt2;
        }
        return cnt1 == 0 && cnt2 == 0;
    }
};

 

下面这种解法使用了哈希表来建立字符和其出现的次数之间的映射,最后直接比较对应的字符出现的次数是否相等即可,参见代码如下:

 

解法二:

class Solution {
public:
    bool judgeCircle(string moves) {
        unordered_map<char, int> m;
        for (char c : moves) ++m[c];
        return m['L'] == m['R'] && m['U'] == m['D'];
    }
};

 

类似题目:

Valid Parentheses

 

参考资料: 

https://discuss.leetcode.com/topic/99256/c-counter-4-lines-solution 

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation